2-Satisfiability (2-SAT) Problem

Boolean Satisfiability Problem

Boolean Satisfiability or simply SAT is the problem of determining if a Boolean formula is satisfiable or unsatisfiable.

Satisfiable : If the Boolean variables can be assigned values such that the formula turns out to be TRUE, then we say that the formula is satisfiable.
Unsatisfiable : If it is not possible to assign such values, then we say that the formula is unsatisfiable.

Examples:

$F = A \wedge \bar{B}$ , is satisfiable, because A = TRUE and B = FALSE makes F = TRUE.
$G = A \wedge \bar{A}$ , is unsatisfiable, because:

$A$ $\bar{A}$ $G$

TRUE FALSE FALSE

FALSE TRUE FALSE

$A$	$\bar{A}$	$G$
TRUE	FALSE	FALSE
FALSE	TRUE	FALSE

Note : Boolean satisfiability problem is NP-complete (For proof, refer <a href="https://en.wikipedia.org/wiki/Cook%E2%80%93Levin_theorem" target="_blank”>Cook’s Theorem).

What is 2-SAT Problem

2-SAT is a special case of Boolean Satisfiability Problem and can be solved
in polynomial time.

To understand this better, first let us see what is Conjunctive Normal Form (CNF) or also known as Product of Sums (POS).
CNF : CNF is a conjunction (AND) of clauses, where every clause is a disjunction (OR).

Now, 2-SAT limits the problem of SAT to only those Boolean formula which are expressed as a CNF with every clause having only 2 terms(also called 2-CNF).

Example: $F = (A_1 \vee B_1) \wedge (A_2 \vee B_2) \wedge (A_3 \vee B_3) \wedge ....... \wedge (A_m \vee B_m)$

Thus, Problem of 2-Satisfiabilty can be stated as:

Given CNF with each clause having only 2 terms, is it possible to assign such values to the variables so that the CNF is TRUE?

Examples:

Input : 
Output : The given expression is satisfiable. 
         (for x1 = FALSE, x2 = TRUE)

Input : 
Output : The given expression is unsatisfiable. 
         (for all possible combinations of x1 and x2)

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach for 2-SAT Problem

For the CNF value to come TRUE, value of every clause should be TRUE. Let one of the clause be $(A \vee B)$ .

$(A \vee B)$ = TRUE

If A = 0, B must be 1 i.e. $(\bar{A} \Rightarrow B)$
If B = 0, A must be 1 i.e. $(\bar{B} \Rightarrow A)$

Thus,

 = TRUE is equivalent to

Now, we can express the CNF as an Implication. So, we create an Implication Graph which has 2 edges for every clause of the CNF.
$(A \vee B)$ is expressed in Implication Graph as $edge(\bar{A} \rightarrow B) \ & edge(\bar{B} \rightarrow A)$
Thus, for a Boolean formula with ‘m’ clauses, we make an Implication Graph with:

2 edges for every clause i.e. ‘2m’ edges.
1 node for every Boolean variable involved in the Boolean formula.

Let’s see one example of Implication Graph.
$F = (x1 \vee x2) \wedge (\bar{x2} \vee x3) \wedge (\bar{x1} \vee \bar{x2}) \wedge (x3 \vee x4) \wedge (\bar{x3} \vee x5) \wedge (\bar{x4} \vee \bar{x5}) \wedge (\bar{x3} \vee x4)$

Note: The implication (if A then B) is equivalent to its contrapositive (if $\bar{B}$ then $\bar{A}$ ).

Now, consider the following cases:

CASE 1: If  exists in the graph
This means 
If X = TRUE,  = TRUE, which is a contradiction.
But if X = FALSE, there are no implication constraints.
Thus, X = FALSE

CASE 2: If  exists in the graph
This means 
If  = TRUE, X = TRUE, which is a contradiction.
But if  = FALSE, there are no implication constraints.
Thus,  = FALSE i.e. X = TRUE

CASE 3: If  both exist in the graph
One edge requires X to be TRUE and the other one requires X to be FALSE.
Thus, there is no possible assignment in such a case.

CONCLUSION: If any two variables $X$ and $\bar{X}$ are on a cycle i.e. $path(\bar{A} \rightarrow B) \& path({B} \rightarrow A)$ both exists, then the CNF is unsatisfiable. Otherwise, there is a possible assignment and the CNF is satisfiable.
Note here that, we use path due to the following property of implication:
If we have $(A \Rightarrow B) \& (B \Rightarrow C), then A \Rightarrow C$
Thus, if we have a path in the Implication Graph, that is pretty much same as having a direct edge.

CONCLUSION FROM IMPLEMENTATION POINT OF VIEW:
If both X and $\bar{X}$ lie in the same SCC (Strongly Connected Component), the CNF is unsatisfiable.
A Strongly Connected Component of a directed graph has nodes such that every node can be reach from every another node in that SCC.
Now, if X and $\bar{X}$ lie on the same SCC, we will definitely have $path(\bar{A} \rightarrow B) \& path({B} \rightarrow A)$ present and hence the conclusion.

Checking of the SCC can be done in O(E+V) using the Kosaraju’s Algorithm

// C++ implementation to find if the given 
// expression is satisfiable using the 
// Kosaraju's Algorithm 
#include <bits/stdc++.h> 
using namespace std; 
  
const int MAX = 100000; 
  
// data structures used to implement Kosaraju's 
// Algorithm. Please refer 
// http://www.geeksforgeeks.org/strongly-connected-components/ 
vector<int> adj[MAX]; 
vector<int> adjInv[MAX]; 
bool visited[MAX]; 
bool visitedInv[MAX]; 
stack<int> s; 
  
// this array will store the SCC that the 
// particular node belongs to 
int scc[MAX]; 
  
// counter maintains the number of the SCC 
int counter = 1; 
  
// adds edges to form the original graph 
void addEdges(int a, int b) 
{ 
    adj[a].push_back(b); 
} 
  
// add edges to form the inverse graph 
void addEdgesInverse(int a, int b) 
{ 
    adjInv[b].push_back(a); 
} 
  
// for STEP 1 of Kosaraju's Algorithm 
void dfsFirst(int u) 
{ 
    if(visited[u]) 
        return; 
  
    visited[u] = 1; 
  
    for (int i=0;i<adj[u].size();i++) 
        dfsFirst(adj[u][i]); 
  
    s.push(u); 
} 
  
// for STEP 2 of Kosaraju's Algorithm 
void dfsSecond(int u) 
{ 
    if(visitedInv[u]) 
        return; 
  
    visitedInv[u] = 1; 
  
    for (int i=0;i<adjInv[u].size();i++) 
        dfsSecond(adjInv[u][i]); 
  
    scc[u] = counter; 
} 
  
// function to check 2-Satisfiability 
void is2Satisfiable(int n, int m, int a[], int b[]) 
{ 
    // adding edges to the graph 
    for(int i=0;i<m;i++) 
    { 
        // variable x is mapped to x 
        // variable -x is mapped to n+x = n-(-x) 
  
        // for a[i] or b[i], addEdges -a[i] -> b[i] 
        // AND -b[i] -> a[i] 
        if (a[i]>0 && b[i]>0) 
        { 
            addEdges(a[i]+n, b[i]); 
            addEdgesInverse(a[i]+n, b[i]); 
            addEdges(b[i]+n, a[i]); 
            addEdgesInverse(b[i]+n, a[i]); 
        } 
  
        else if (a[i]>0 && b[i]<0) 
        { 
            addEdges(a[i]+n, n-b[i]); 
            addEdgesInverse(a[i]+n, n-b[i]); 
            addEdges(-b[i], a[i]); 
            addEdgesInverse(-b[i], a[i]); 
        } 
  
        else if (a[i]<0 && b[i]>0) 
        { 
            addEdges(-a[i], b[i]); 
            addEdgesInverse(-a[i], b[i]); 
            addEdges(b[i]+n, n-a[i]); 
            addEdgesInverse(b[i]+n, n-a[i]); 
        } 
  
        else
        { 
            addEdges(-a[i], n-b[i]); 
            addEdgesInverse(-a[i], n-b[i]); 
            addEdges(-b[i], n-a[i]); 
            addEdgesInverse(-b[i], n-a[i]); 
        } 
    } 
  
    // STEP 1 of Kosaraju's Algorithm which 
    // traverses the original graph 
    for (int i=1;i<=2*n;i++) 
        if (!visited[i]) 
            dfsFirst(i); 
  
    // STEP 2 pf Kosaraju's Algorithm which 
    // traverses the inverse graph. After this, 
    // array scc[] stores the corresponding value 
    while (!s.empty()) 
    { 
        int n = s.top(); 
        s.pop(); 
  
        if (!visitedInv[n]) 
        { 
            dfsSecond(n); 
            counter++; 
        } 
    } 
  
    for (int i=1;i<=n;i++) 
    { 
        // for any 2 vairable x and -x lie in 
        // same SCC 
        if(scc[i]==scc[i+n]) 
        { 
            cout << "The given expression "
                 "is unsatisfiable." << endl; 
            return; 
        } 
    } 
  
    // no such variables x and -x exist which lie 
    // in same SCC 
    cout << "The given expression is satisfiable."
         << endl; 
    return; 
} 
  
//  Driver function to test above functions 
int main() 
{ 
    // n is the number of variables 
    // 2n is the total number of nodes 
    // m is the number of clauses 
    int n = 5, m = 7; 
  
    // each clause is of the form a or b 
    // for m clauses, we have a[m], b[m] 
    // representing a[i] or b[i] 
  
    // Note: 
    // 1 <= x <= N for an uncomplemented variable x 
    // -N <= x <= -1 for a complemented variable x 
    // -x is the complement of a variable x 
  
    // The CNF being handled is: 
    // '+' implies 'OR' and '*' implies 'AND' 
    // (x1+x2)*(x2’+x3)*(x1’+x2’)*(x3+x4)*(x3’+x5)* 
    // (x4’+x5’)*(x3’+x4) 
    int a[] = {1, -2, -1, 3, -3, -4, -3}; 
    int b[] = {2, 3, -2, 4, 5, -5, 4}; 
  
    // We have considered the same example for which 
    // Implication Graph was made 
    is2Satisfiable(n, m, a, b); 
  
    return 0; 
} 

Output:

The given expression is satisfiable.

More Test Cases:

Input : n = 2, m = 3
        a[] = {1, 2, -1}
        b[] = {2, -1, -2}
Output : The given expression is satisfiable.

Input : n = 2, m = 4
        a[] = {1, -1, 1, -1}
        b[] = {2, 2, -2, -2}
Output : The given expression is unsatisfiable.

This article is contributed by Aanya Jindal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Boolean Satisfiability Problem

What is 2-SAT Problem

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach for 2-SAT Problem

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